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Math · Calculus I · Worked example

Estimate a square root with a tangent line

Use a linear approximation to estimate √4.1.

f⁡(x)=x,a=4

Choose f and a

Let f(x) = √x and a = 4, a nearby point where the square root is exact: f(4) = 2.

f⁡(x)=x,f⁡(4)=2

Find the slope at a

f′(x) = 1/(2√x), so f′(4) = 1/4.

f⁡′(x)=12x,f⁡′(4)=14

Write the linearization

L(x) = f(4) + f′(4)(x − 4) = 2 + (x − 4)/4.

L⁢(x)=2+14⁢(x−4)

Evaluate near a

L(4.1) = 2 + (1/4)(0.1) = 2.025.

L⁢(4.1)=2+14⁢(0.1)=2.025

Compare with the true value

√4.1 = 2.024846…, so the error is about 0.00015. The estimate is slightly high because √x is concave down: its graph bends below the tangent line.

Result

√4.1 ≈ 2.025. The true value is 2.02485 to five decimal places.

4.1≈2.025

Your turn

Use the linearization of f(x) = ∛x at a = 27 to estimate ∛27.5.

Show the answer and explanation

∛27.5 ≈ 3.0185.

Write f(x) = x^(1/3). Then f(27) = 3 and f′(x) = (1/3)x^(−2/3), so f′(27) = 1/(3 · 9) = 1/27. L(27.5) = 3 + 0.5/27 = 3.0185. The true value is 3.01841, so the estimate is slightly high, as it is for √x.

f⁡(x)=x13f⁡′(x)=13x−23f⁡′(27)=127L⁢(x)=3+127⁢(x−27)L⁢(27.5)≈3.0185

Keep exploring

Open the rows in Math and add L(3.9) = 1.975: the same tangent line estimates √3.9, and the checker confirms the value.

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